guess-number-higher-or-lower-ii.py

# Time:  O(n^2)
# Space: O(n^2)

# We are playing the Guess Game. The game is as follows:
#
# I pick a number from 1 to n. You have to guess which number I picked.
#
# Every time you guess wrong, I'll tell you whether the number I picked is higher or lower.
#
# However, when you guess a particular number x, and you guess wrong,
# you pay $x. You win the game when you guess the number I picked.
#
# Example:
#
# n = 10, I pick 8.
#
# First round:  You guess 5, I tell you that it's higher. You pay $5.
# Second round: You guess 7, I tell you that it's higher. You pay $7.
# Third round:  You guess 9, I tell you that it's lower. You pay $9.
#
# Game over. 8 is the number I picked.
#
# You end up paying $5 + $7 + $9 = $21.
# Given a particular n >= 1, find out how much money you need to have to guarantee a win.
#
# Hint:
#
# The best strategy to play the game is to minimize the maximum loss
# you could possibly face. Another strategy is to minimize the expected loss.
# Here, we are interested in the first scenario.
# Take a small example (n = 3). What do you end up paying in the worst case?
# Check out this article if you're still stuck.
# The purely recursive implementation of minimax would be worthless
# for even a small n. You MUST use dynamic programming.
# As a follow-up, how would you modify your code to solve the problem of
# minimizing the expected loss, instead of the worst-case loss?

class Solution(object):
    def getMoneyAmount(self, n):
        """
        :type n: int
        :rtype: int
        """
        pay = [[0] * n for _ in xrange(n+1)]
        for i in reversed(xrange(n)):
            for j in xrange(i+1, n):
                pay[i][j] = min(k+1 + max(pay[i][k-1], pay[k+1][j]) \
                                for k in xrange(i, j+1))
        return pay[0][n-1]